
#1
Feb1913, 07:41 AM

P: 13

i need to find a series of numbers up to n that will add up to kn
x1 + x2 + x3+ ... + n = kn where k is a constant. this is part of a long complex problem once this sum is found it will finally be solved. 



#2
Feb1913, 08:18 AM

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How about ##x_1=x_2=\cdots =x_{n1}=x_n = k##? 



#3
Feb1913, 04:45 PM

P: 13

no, i need the last number to be n.
some how i need to find a set of numbers when added together all i get is a multiple of the last number. 



#4
Feb1913, 05:38 PM

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Summation of a set = kn? 



#5
Feb1913, 07:49 PM

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... which means that you have to be more specific about your problem if you are to get sensible answers. I'm guessing the x's can't be just any old numbers for example, and k is not supposed to be a specific integer? But I'm guessing  don't make people guess. 



#6
Feb2013, 01:42 PM

P: 13

yeah, i should've added more restrictions.
well the numbers must increase as we go on, for example: 1+2+3+...+n=n(n+1)/2 right, but this is of second order i want to add numbers that increase up to n, and their sum is of order 1. 



#7
Feb2013, 08:32 PM

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#8
Feb2113, 01:46 AM

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If this is a problem that has been supplied to you, then please provide the exact text of the problem and the context.
I'll add that we don't know that the numbers have to be integers of if they can be real, and so on. eg. ##\sum_{i=1}^n = n(n+1)/2 = kn \text{ if } k=(n+1)/2## But what do you mean "the sum is of order 1"? You mean that kn cannot be quadratic in n? (k cannot depend on n?) Maybe you are thinking more like: x1=1, x2=2, x3=3=n, then 1+2+3=6=2n so k=2 that the idea? 



#9
Feb2213, 10:43 AM

P: 13

Well the problem is not about this sum, if you find a series like the one described the problem will be solved.
i have submitted the following answer: 2+4+8+16+32+64+...+2^k = 2(1+2+4+8+...+2^(k1) = 2(2^k 1)/21 = 2(2^k 1) but 2^k = n k = log (base 2) n right, then substitute in the sum: 2(n1) which is linear that is (of order one). the idea of the problem is that i want to get to n with the least sum possible, if i went with the quadratic equation that would have taken too much time O(n^2), so this sum will save so much time O(n) it's an algorithm problem. 


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